1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find ##v_o## in the network

  1. Sep 20, 2014 #1

    Zondrina

    User Avatar
    Homework Helper

    I was able to do most of the exercises, but this one gave me some trouble. I want to find ##V_o##.

    Screen Shot 2014-09-20 at 1.48.52 PM.png

    My gut is telling me to use KCL, and applying it to the top left node above the 6k resistor yields:

    ##\frac{V_o}{2000} + 2 \times 10^{-3} - \frac{V_s}{6000} - \frac{V_s}{3000} = 0##

    I am now somewhat unsure how to proceed. I know the potential across the branches is ##V_s##.
     
  2. jcsd
  3. Sep 20, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    Write another equation for the branch with the two resistors and Vo. The equation should relate Vo to the voltage Vs (the voltage of the node). Then you'll have something to replace Vo in your node equation, leaving you with just one unknown.
     
  4. Sep 20, 2014 #3

    Zondrina

    User Avatar
    Homework Helper

    I suspected I was trying to relate ##V_s## and ##V_o##, but I can't see it for some reason. Probably due to hunger. Applying KVL didn't seem to give me anything useful, so I'm still a bit unsure what to do.
     
  5. Sep 20, 2014 #4

    Zondrina

    User Avatar
    Homework Helper

    The only equation I can see is: ##V_s = V_o + V_x## where ##V_x## is the voltage across the 1k.
     
  6. Sep 20, 2014 #5

    gneill

    User Avatar

    Staff: Mentor

    The potential at the top node is ##V_s##. That branch with ##V_o## looks like a voltage divider to me... ;)
     
  7. Sep 20, 2014 #6

    Zondrina

    User Avatar
    Homework Helper

    I can't believe I forgot entirely about voltage dividers.

    So I guess I could just say that ##V_o = [\frac{2k}{1k + 2k}]V_s = \frac{2}{3} V_s##

    Subbing this into the KCL equation gives ##V_s = 12 V##, which then subbing back gives ##V_o = 8 V##.

    I'm going to go eat now, thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find ##v_o## in the network
Loading...